Question

If $\left|\begin{array}{ccc}{x}^n& x^{n+2}& x^{n+3}\\ y^n& y^{n+2}& y^{n+3}\\ z^n& z^{n+2}& z^{n+3}\end{array}\right|=(y-z)(z-x)(x-y)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$ then n is equal to :

• A. 2
• B. -2
• C. -1
• D. 1

Answer 1

Answer: (C)

Given: $\left|\begin{array}{ccc}{x}^n& x^{n+2}& x^{n+3}\\ y^n& y^{n+2}& y^{n+3}\\ z^n& z^{n+2}& z^{n+3}\end{array}\right|$
$=\left(x-y\right)\left(y-z\right)\left(z-x\right)\left[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right]$
Let $A=\left|\begin{array}{ccc}{x}^n& x^{n+2}& x^{n+3}\\ y^n& y^{n+2}& y^{n+3}\\ z^n& z^{n+2}& z^{n+3}\end{array}\right|$
Take $x^n,y^n$ and $z^n$ common from the rows $R_1,R_2$ and $R_3$ respectively. $A=x^n{y}^n{z}^n\left|\begin{array}{ccc}1& x^2& x^3\\ 1& y^2& y^3\\ 1& z^2& z^3\end{array}\right|$
Operate $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$
$A=x^n{y}^n{z}^n\left|\begin{array}{ccc}1& x^2& x^3\\ 0& y^2-x^2& y^3-x^3\\ 0& z^2-x^2& z^3-x^3\end{array}\right|$
Take (y - x) (z - x) common from the rows $R_2$ and $R_3$ respectively:
Then we get $A=x^n{y}^n{z}^n(y-x)(z-x)$
$\left|\begin{array}{ccc}1& x^2& x^3\\ 0& y+x& x^2+y^2+xy\\ 0& z+x& z^2+x^2+zx\end{array}\right|$
$=x^n{y}^n{z}^n\left(y-x\right)\left(z-x\right)\left[y{z}^2+x^{2}y+xyz+x{z}^2+x^3+x^{2}z-x^{2}z-y^{2}z-xyz-x^3-x{y}^2-x^{2}y\right]$
$=x^n{y}^n{z}^n\left(x-y\right)\left(y-z\right)\left(z-x\right)\left[xy+yz+zx\right]$
$=\left[x^{n+1}{y}^{n+1}{z}^n+x^n{y}^{n+1}{z}^{n+1}+x^{n+1}{y}^n{z}^{n+1}\right]$
$\left(x-y\right)\left(y-z\right)\left(z-x\right)$
$=x^{n+1}.y^{n+1}.z^{n+1}\left(x-y\right)\left(y-z\right)\left(z-x\right)$
$\left[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right]$
But given
$A=\left(y-z\right)\left(z-x\right)\left(x-y\right)\left[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right]$
So, $x^{n+1}{y}^{n+1}{z}^{n+1}=1=x^0{y}^0{z}^0$
$\Rightarrow n+1=0\Rightarrow n=-1$.